Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e4 + 5;
int n, a[maxn], b[maxn];
pair<int, int> opa[maxn * 20], opb[maxn * 20];
void Swap(int *a, int x, int y, int &cnt, pair<int, int> *op) {
  if (x == y) return;
  op[cnt++] = make_pair(x, y);
  op[cnt++] = make_pair(y, x);
  op[cnt++] = make_pair(x, y);
  swap(a[x], a[y]);
}
void Gauss(int *a, int &cnt, pair<int, int> *op) {
  int r = 0;
  for (int i = 30; i >= 0; --i) {
    for (int j = r; j < n; ++j)
      if (a[j] & (1 << i)) {
        Swap(a, r, j, cnt, op);
        for (int k = 0; k < n; ++k)
          if (k != r && a[k] & (1 << i)) {
            a[k] ^= a[r];
            op[cnt++] = make_pair(k, r);
          }
        ++r;
        break;
      }
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) scanf("%d", a + i);
  for (int i = 0; i < n; ++i) scanf("%d", b + i);
  int ca = 0, cb = 0;
  Gauss(a, ca, opa);
  Gauss(b, cb, opb);
  for (int i = 0; i < 40 && i < n; i++) {
    int p = 30;
    if (a[i] == 0) continue;
    while ((a[i] & (1 << p)) == 0) p--;
    for (int j = 0; j < 40 && j < n; j++)
      if ((a[j] ^ b[j]) & (1 << p)) {
        opa[ca++] = make_pair(j, i);
        a[j] ^= a[i];
      }
  }
  for (int i = 0; i < n; ++i)
    if (a[i] != b[i]) return puts("-1");
  printf("%d\n", ca + cb);
  for (int i = 0; i < ca; ++i)
    printf("%d %d\n", opa[i].first + 1, opa[i].second + 1);
  for (int i = cb - 1; i >= 0; --i)
    printf("%d %d\n", opb[i].first + 1, opb[i].second + 1);
  return 0;
}